Q16 / 150 · MHT CET 2026 (15 April Shift 1) · Chemistry
Calculate the van't Hoff factor of a centimolar solution of potassium ferrocyanide if it is $60\%$ dissociated at $300$ K.
- Correct answerA$3.4$
- B$3.9$
- C$3.1$
- D$3.7$
Solution
The chemical formula of potassium ferrocyanide is $K_4[Fe(CN)_6]$.
The dissociation reaction is:
$K_4[Fe(CN)_6] \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$
The number of ions produced per formula unit, $n = 5$.
The degree of dissociation, $\alpha = 60\% = 0.60$.
The van't Hoff factor $i$ is given by:
$i = 1 + (n - 1)\alpha$
Substituting the values:
$i = 1 + (5 - 1) \times 0.60$
$i = 1 + 4 \times 0.60$
$i = 1 + 2.4 = 3.4$
Answer: $3.4$
The dissociation reaction is:
$K_4[Fe(CN)_6] \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$
The number of ions produced per formula unit, $n = 5$.
The degree of dissociation, $\alpha = 60\% = 0.60$.
The van't Hoff factor $i$ is given by:
$i = 1 + (n - 1)\alpha$
Substituting the values:
$i = 1 + (5 - 1) \times 0.60$
$i = 1 + 4 \times 0.60$
$i = 1 + 2.4 = 3.4$
Answer: $3.4$
Hence, the correct answer is option: 'A'.